How many faces does an icosahedron have? | Reference.com How many faces does an icosahedron have? A: Quick Answer An icosahedron is a regular polyhedron with 20 faces, all of which are equilateral triangles. An icosahedron has 30 edges and 12 vertices. Five equilateral triangles meet at each vertex. Full Answer An icosahedron is an example of a Platonic solid, which is a polyhedron that is regular, convex, made of congruent faces of regular polygons, and has the same number of faces meeting at each vertex. Only five known shapes meet this criteria. The other Platonic solids include the tetrahedron with four triangular faces, the cube, or hexahedron, with six square faces, the octahedron with eight triangular faces and the dodecahedron with twelve pentagonal faces. Icosahedron -- from Wolfram MathWorld ), 18-gonal prism, 19-gonal pyramid, regular icosahedron, and rhombic icosahedron .

"The" icosahedron (more properly called the regular icosahedron) is the regular polyhedron and Platonic solid illustrated above having 12 polyhedron vertices , 30 polyhedron edges , and 20 equivalent equilateral triangle faces, . The regular icosahedron is also uniform polyhedron and Wenninger model and Wythoff symbol . Coxeter et al. (1999) have shown that there are 58 icosahedron stellations (giving a total of 59 solids when the icosahedron itself is included). Two icosahedra constructed in origami are illustrated above (Gurkewitz and Arnstein 1995, p. 53). This construction uses 30 triangle edge modules, each made from a single sheet of origami paper. Two icosahedra appears as polyhedral "stars" in M. C. Escher's 1948 wood engraving "Stars" (Forty 2003, Plate 43).

There are 43380 distinct nets for the icosahedron, the same number as for the dodecahedron (Bouzette and Vandamme, Hippenmeyer 1979, Buekenhout and Parker 1998). The icosahedron has the icosahedral group of symmetries. The connectivity of the vertices is given by the icosahedral graph . The dual polyhedron of an icosahedron with unit edge lengths is the dodecahedron with edge lengths , where is the golden ratio . As a result, the centers of the faces of an icosahedron form a dodecahedron , and vice versa, illustrated above (Steinhaus 1999, pp. 199-201). In particular, fifteen golden rectangles span the interior of the icosahedron. These rectangles have 30 edges, and each edge pairs up with its opposite edge to form a golden rectangle There are 59 distinct icosahedra when each triangle is colored differently (Coxeter 1969).

More general question of polyhedron coloring can be addressed using the Pólya enumeration theorem . Taken eight at a time, the centers of the faces of an icosahedron comprise the vertices of a cube . This leads to the beautiful cube 5-compound and is the basis for Jessen's orthogonal icosahedron . A plane perpendicular to a axis of an icosahedron cuts the solid in a regular decagonal cross section (Holden 1991, pp. 24-25). The long diagonals of the faces of the rhombic triacontahedron give the edges of an icosahedron (Steinhaus 1999, pp. 209-210). The following table gives polyhedra which can be constructed by cumulation of an icosahedron by pyramids of given heights . A construction for an icosahedron with side length places the end vertices at ) and the central vertices around two staggered circles of radii and heights . By a suitable rotation, the polyhedron vertices of an icosahedron of side length 2 can also be placed at , , and , where is the golden ratio .

These points divide the polyhedron edges of an octahedron into segments with lengths in the ratio . Another orientation of the icosahedron places two opposite triangular faces in an orientation parallel to the -plane. In this orientation, the distance from the top plane to the triangle of vertices below it is , equal to the circumradius of a face. The circumradius of (1) To derive the volume of an icosahedron having edge length , consider the orientation so that two polyhedron vertices are oriented on top and bottom. The vertical distance between the top and bottom pentagonal dipyramids is then given by (2) The Icosahedron The Icosahedron The icosahedron has 12 vertices, 20 faces and 30 sides. It is one of the most interesting and useful of all polyhedra. Buckminster Fuller based his designs of geodesic domes around the icosahedron. The icosahedron is built around the pentagon and the golden section. At first glance this may seem absurd, since every face of the icosahedron is an equilateral triangle.

It turns out, however, that the triangular faces of the icosahedron result from its pentagonal nature. First, weï¿½ll display 3 views of this polyhedron: Figure 1 Figure 3 Figure 1 shows 2 of the internal pentagons of the icosahedron, LGFDC, and ABKIH. Figure 2 helps to show that the icosahedron is actually an interlocking series of pentagons. Notice the exterior pentagons at CDFHL, CEHIK, EFHIL, and CDELK. Figure 3 shows a 2 dimensional 'shadow' of the icosa from the top down. You can see that the outer edges form a perfect decagon, formed by the 2 pentagons CDFGL and ABKIH. In fact, every vertex of the icosahedron is the vertex of a pentagon. OK, lets go through the usual analysis and then get on to the interesting stuff! First. lets calculate the volume of the icosa. Using the pyramid method we have 20 equilateral triangular faces which serve as the base to a pyramid whose topmost point is the centroid of the icosa.

Figure 4 Some icosahedron faces with midpoints. All 20 faces will be connected to O to form 20 pyramids. Figure 5 1 icosahedral pyramid The volume of each pyramid is 1/3 * area of base * pyramid height. The area of the base is the area of the equilateral triangle ADB. The height of the pyramid is OM. From The Equilateral Triangle we know the area is All vertices of the icosahedron (as with all 5 of the regular solids) lie upon the surface of a sphere that encloses it. The radius of the circumsphere is O to any vertex, in this case, r = OA = OB = OD = 1. The height of the pyramid is h = OM. In order to find h we need to recognize that any of the triangles OMA, OMB, OMD are right. This is because OM is perpendicular to the plane of triangle ABD by construction. AB = BD = AD = side of icosahedron = s.

Lets work with triangle OMD: Figure 6 We know, from The Equilateral Triangle that DM = In order to find h, we need to find OD = r, in terms of the side s of the icosahedron. To do that, we have to recognize one of the basic geometric properties of the icosahedron. Take a look at Figure 1. DI, BG and FK are all diameters of the enclosing sphere around the icosahedron. Notice rectangle BFGK and notice that BG, FK are both diagonals of it. Now notice that both FB and GK are both diagonals of the 2 pentagonal planes marked in gray. We know from Composition of the Pentagon that the diagonal of a pentagon is * side of pentagon. * side of icosahedron, since each side of the pentagon is a side of the icosahedron. FB = * s Figure 7 FK is the diameter of the enclosing sphere around the icosahedron. OF = OK is the radius r, which we are now trying to find. FG = BK is the side s of the icosahedron. We have found that FB = GK = * s .

There are 20 pyramids, 1 for each face so = = 2.181694991sï¿½. What is the surface area of the icosahedron? It is just the area of 1 face * 20 faces. The area of each face is, from above, . So We have already noted the relationship between the radius of the enclosing sphere and the side of the icosahedron: r = .951056517s, s = 1.051462224r. The side or edge of the icosahedron is slightly larger than the radius. What is the central angle of the icosahedron? The central angle, DOB, can be seen clearly from Figure 5, and we diagram it in Figure 8 below. Figure 8 s. DB is the side of the icosahedron, s. Dihedral angle = 138.1896852ï¿½ Now lets figure out the distances from the centroid of the icosahedron to any vertex, to any mid-face, and to any mid-side.

We already have the first two (see Figure 10). OD = OA = OB = r = OM = h = Now we need to find, for example, OX. Figure 10 If we lay a 3D model of the icosahedron on one of its sides, we can see that a line through the centroid O is perpendicular to that side. So the triangle OXB, for example, is right. We know AB = s, so XB = (1/2)s. OB = r, so OXï¿½ = OBï¿½ - BXï¿½ = distance from centroid to mid-face = = 0.755761314s. distance from centroid to mid-side = = 0.809016995s. distance from centroid to a vertex = = 0.951056517s. Comparing distances, we have .755761314s, .809016995s, .951056517s. Now lets get to the interesting stuff!

Going back to figure 1, we can see that the icosahedron is composed of interlocking pentagonal 'caps.' Look at D-ABCEF and I-GHJKL to see this more clearly. Of course , EVERY vertex of the icosahedron is the top of a pentagonal cap, not just D and I. Letï¿½s analyze this cap: Figure 12 Triangle DQF is right, by construction. D is directly over the point Q in the icosahedron. How far off the plane of pentagon AFECB is D? In other words, what is the distance DQ? First, realize there is a circle around pentagon AFECB, even though I havenï¿½t draw it here. QF is the radius of that circle. We know from Construction of the Pentagon, Part 2 that this radius r = FQ = Here, s = the side of the icosahedron. DF is a side of the icosahedron, so DF = s. Therefore we write DQï¿½ = DFï¿½ - FQï¿½ = DQ = Hmmm, this is looking interesting.

Lets compare FQ to DQ. FQ / DQ = In order to form equilateral triangles with D and the other vertices of the pentagon, D has to be raised off the pentagonal plane AFECB by the division of the radius of the pentagon (FQ) in Mean and Extreme Ratio. The triangle DQF is therefore a Phi based triangle, specifically, a triangle. From The Phi Triangle we know that DFQ = 31.71747441ï¿½. Is this surprising? It was to me! I didn't expect something that was composed entirely of equilateral triangles to have any relationship to . An equilateral triangle is geometry. Figure 13 Here we have Figure 1 basically, with the mid-face points of the 2 pentagons marked off as Q and Z. We have already seen that DQ is with respect to the side of the icosahedron. That means IZ = s as well, since IZ = DQ. What about OQ = OZ? and QZ? How do all of these distances relate to the diameter of the enclosing sphere, DI?

Remember that the distance DI = FK = BG, etc., is the diagonal of any of the rectangles of which the icosa is composed. One of these rectangles is clearly visible in Figure 12 as BFGK. We know this is a rectangle because it is the diagonal of the pentagon ABCEF. In fact, if you place 3 of these rectangles perpendicular to each other, the 12 corners of the 3 rectangles are the vertices of the icosahedron! FB = Figure 14. Central axis (diameter) relationships diameter = DI (Available in the book). DZ is divided in EMR by Q, IQ is divided in EMR by Z. All of these relationships come from the pentagon! On the outside of the icosahedron, we see equilateral triangles. But the guts of this polyhedron comes from pentagonal relationships. The equilateral triangles come about from the lifting of the pentagonal 'cap' vertex off the pentagonal plane. There is now no question that the basis for the construction of the icosahedron is the pentagon. Or is there?

Take a look at this view of the icosahedron: Figure 15 As you can see, the 2 pentagonal planes in the middle have magically disappeared and become equilateral triangles! All we have done is placed the icosahedron on one of its faces, dropping it from 30ï¿½ along the x axis and 30ï¿½ back along the negative y axis. It seems that all of our work is wrong, except: The sides of the equilateral triangles EHB and JGC are all diagonals of pentagons! EH is a diagonal of pentagon FEJIH, EB is a diagonal of pentagon DEJKB, and HB is a diagonal of pentagon AHIKB. The equilateral triangle faces of the icosahedron are a by-product of how the pentagons interlock. This is shown in Figure 16 below: Figure 16 -- attempting to show the individual pentagons of the icosahedron Because the pentagons are interlocking, there is duplication of colors, which makes the individual pentagons a bit difficult to identify.

Begin with the two pentagons BDFHJ, in green, and CEGIK, in black. Insert pentagon AJKCD, 4 of which sides are in blue. Insert pentagon ABKIH, 4 of which sides are marked in red. Insert pentagon BCLIJ, 4 of which sides are marked in orange. Insert pentagon ADEGH, 2 of which sides are marked in magenta. Insert pentagon AFECB, 2 of which sides are marked in gold. Insert pentagon CDFGL, 2 of which sides are marked in cyan (sky blue) The side KL in white is part of the pentagon KLGHJ. The side LE is part of the pentagon ELIHF.

Icosaahedron Reference Tables (included in the book) Return to Geometry Home Page ï¿½ï¿½ï¿½ï¿½ The Big Picture Home Spinning Icosahedron It has 12 Vertices (corner points) and at each vertex 5 edges meet It is one of the Platonic Solids Volume and Surface Area Volume = 5×(3+√5)/12 × (Edge Length)3 Surface Area = 5×√3 × (Edge Length)2 It is called an icosahedron because it is a polyhedron that has 20 faces (from Greek icosa- meaning 20) When we have more than one icosahedron they are called icosahedra When we say "icosahedron" we often mean "regular icosahedron" (in other words all faces are the same size and shape), but it doesn't have to be - this is also an icosahedron, even though all faces are not the same. 20-Sided Dice? Yes! An icosahedron that has 20 equal faces has an equal chance of landing on any face.

In fact, you can make fair dice out of all of the Platonic Solids. The outer shell of this virus (the Human papilloma virus) is like an icosahedron Courtesy of VirusWolrd at virology.wisc.edu cut out the shape and glue it together. Icosahedron | Math Wiki | Fandom powered by Wikia of a regular icosahedron of edge length are: Cartesian coordinates The following Cartesian coordinates define the vertices of an icosahedron with edge-length 2, centered at the origin: where is the golden ratio (also written τ). Note that these vertices form five sets of three mutually centered, mutually orthogonal golden rectangles . The 12 edges of a regular octahedron can be partitioned in the golden ratio so that the resulting vertices define a regular icosahedron. This is done by first placing vectors along the octahedron's edges such that each face is bounded by a cycle, then similarly partitioning each edge into the golden mean along the direction of its vector.

The five octahedra defining any given icosahedron form a regular polyhedral compound , as do the two icosahedra that can be defined in this way from any given octahedron. Construction by a system of equiangular lines The following construction of the icoshaedron avoids tedious computations in the number field necessary in more elementary approaches. Existence of the icosahedron amounts to the existence of six equiangular lines in . Indeed, intersecting such a system of equiangular lines with an Euclidean sphere centered at their common intersection yields the twelve vertices of a regular icosahedron as can easily be checked. Conversely, supposing the existence of a regular icosahedron, lines defined by its six pairs of opposite vertices form an equiangular system. In order to construct such an equiangular system, we start with the matrix of square size encodes the angles, acute for a positive entry, obtuse otherwise, between five cyclically consecutive vertices adjacent to a first vertex of an icosahedron centered at the origin.) A straightforward computation yields yields thus the twelve vertices of the icosahedron.

A second straightforward construction of the icosahedron uses representation theory of the alternating group acting by direct isometries on the icosahedron. Stellations According to specific rules defined in the book The fifty nine icosahedra , 59 stellations were identified for the regular icosahedron. The first form is the icosahedron itself. One is a regular Kepler-Poinsot solid . Three are regular compound polyhedra . [1] 21 of 59 stellations The faces of the icosahedron extended outwards as planes intersect, defining regions in space as shown by this stellation diagram of the intersections in a single plane. Geometric relations Icosahedron as a snub tetrahedron . Icosahedron as an alternated truncated octahedron. There are distortions of the icosahedron that, while no longer regular, are nevertheless vertex-uniform. These are invariant under the same rotations as the tetrahedron , and are somewhat analogous to the snub cube and snub dodecahedron , including some forms which are chiral and some with Th-symmetry, i.e.

have different planes of symmetry from the tetrahedron. The icosahedron has a large number of stellations , including one of the Kepler-Poinsot polyhedra and some of the regular compounds, which could be discussed here. The icosahedron is unique among the Platonic solids in possessing a dihedral angle not less than 120°. Its dihedral angle is approximately 138.19°. Thus, just as hexagons have angles not less than 120° and cannot be used as the faces of a convex regular polyhedron because such a construction would not meet the requirement that at least three faces meet at a vertex and leave a positive defect for folding in three dimensions, icosahedra cannot be used as the cells of a convex regular polychoron because, similarly, at least three cells must meet at an edge and leave a positive defect for folding in four dimensions (in general for a convex polytope in dimensions, at least three facets must meet at a peak and leave a positive defect for folding in n-space).

However, when combined with suitable cells having smaller dihedral angles, icosahedra can be used as cells in semi-regular polychora (for example the snub 24-cell ), just as hexagons can be used as faces in semi-regular polyhedra (for example the truncated icosahedron ). Finally, non-convex polytopes do not carry the same strict requirements as convex polytopes, and icosahedra are indeed the cells of the icosahedral 120-cell , one of the ten non-convex regular polychora. An icosahedron can also be called a gyroelongated pentagonal bipyramid . It can be decomposed into a gyroelongated pentagonal pyramid and a pentagonal pyramid or into a pentagonal antiprism and two equal pentagonal pyramids . The icosahedron can also be called a snub tetrahedron, as snubification of a regular tetrahedron gives a regular icosahedron.

Alternatively, using the nomenclature for snub polyhedra that refers to a snub cube as a snub cuboctahedron (cuboctahedron = rectified cube) and a snub dodecahedron as a snub icosidodecahedron (icosidodecahedron = rectified dodecahedron), one may call the icosahedron the snub octahedron (octahedron = rectified tetrahedron). A rectified icosahedron forms an icosidodecahedron . Icosahedron vs dodecahedron When an icosahedron is inscribed in a sphere , it occupies less of the sphere's volume (60.54%) than a dodecahedron inscribed in the same sphere (66.49%). Also, as these are duals, it is possible to transform one into the other(See below). Icosahedron File:Herpes simpex virus.jpg Many viruses , e.g. herpes virus, have the shape of an icosahedron.

Viral structures are built of repeated identical protein subunits and the icosahedron is the easiest shape to assemble using these subunits. A regular polyhedron is used because it can be built from a single basic unit protein used over and over again; this saves space in the viral genome . In 1904, Ernst Haeckel described a number of species of Radiolaria , including Circogonia icosahedra, whose skeleton is shaped like a regular icosahedron. A copy of Haeckel's illustration for this radiolarian appears in the article on regular polyhedra . Twenty-sided die. In some roleplaying games , the twenty-sided die (for short, d20 ) is used in determining success or failure of an action. This die is in the form of a regular icosahedron. It may be numbered from "0" to "9" twice (in which form it usually serves as a ten-sided die , or d10 ), but most modern versions are labeled from "1" to "20". See d20 System .

An icosahedron is the three-dimensional game board for Icosagame , formerly known as the Ico Crystal Game. An icosahedron is used in the board game Scattergories to choose a letter of the alphabet. Six little-used letters, such as X, Q, and Z, are omitted. Inside a Magic 8-Ball , various answers to yes-no questions are printed on a regular icosahedron. The icosahedron displayed in a functional form is seen in the Sol de la Flor light shade. The rosette formed by the overlapping pieces show a resemblance to the Frangipani flower. If each edge of an icosahedron is replaced by a one ohm resistor , the resistance between opposite vertices is 0.5 ohms, and that between adjacent vertices 11/30 ohms. [2] The symmetry group of the icosahedron is isomorphic to the alternating group on five letters. This nonabelian simple group is the only nontrivial normal subgroup of the symmetric group on five letters.

Since the Galois group of the general quintic equation is isomorphic to the symmetric group on five letters, and the fact that the icosahedral group is simple and nonabelian means that quintic equations need not have a solution in radicals. The proof of the Abel-Ruffini theorem uses this simple fact, and Felix Klein wrote a book that made use of the theory of icosahedral symmetries to derive an analytical solution to the general quintic equation. See also polygons and polyhedra : paul scott : icosahedron Circumradius Model making Making a model of the icosahedron is a little fiddly with 20 faces, but it is not difficult to make. Go to this construction page for some face templates. A net for the regular icosahedron is given at right. It is known that the icosahedron (like the dodecahedron) has 43380 distinct nets! Projections and sections The icosahedron has a number of interesting properties. We shall use our applet to investigate some of these.

 As with the dodecahedron, the projections of the icosahedron are less interesting than those of some of the other Platonic solids but you should be able to obtain a {6}, a non-regular octagon, and a {10}. The cross-sections make us work rather harder. Working from a model (if you have one) is probably easier than working from the applet.  You can try to find the sequence of cross-sections as a plane moves through the dodecahedron cutting it from vertex to vertex, edge to edge, and face to face. Alternatively, see if you can find a cross section which is a {3}, a {5}, a {10}. Note where these shapes occur. Inscribed polyhedra As we did with the dodecahedron, we now ask whether it is possible to inscribe a tetrahedron, a cube, or an octahedron nicely inside a dodecahedron.  (a) The simplest is the octahedron. How many vertices does the octahedron have? Can you find six nicely related edges of the icosahedron?

Check your ideas from one of the above figures, or using the applet . Briefly describe your solution below. (b) Next investigate the cube. Use the applet to find four points which will serve as vertices of a face. Possible suitable points might include icosahedral vertices, midpoints of edges, and centres of faces. Briefly describe your solution below. Are you certain it is a solution?! (c) It should now be trivial to find a regular tetrahedron inscribed in the icosahedron. Why is this? Describe your solution below. The diagrams below show that each of these inscribed polyhedra is possible. For the octahedron on the right, we observe that we can choose three parallel pairs of edges of the icosahedron which are mutually perpendicular. The midpoints of these six edges form the vertices of a regular inscribed octahedron. [Are the three pairs of edges parallel and mutually perpendicular? Consider one of the pentagonal pyramids forming the icosahedron, and let e denote one of the edges from the vertex. The plane of symmetry of the pyramid containing e passes through the altitude of the opposite triangular face, cutting the base f of that triangle perpendicularly.

Thus edges e and f are perpendicular. Our result follows by applying this argument to each of the three pairs of edges in turn.] In the case of the cube, we choose as vertices eight triangular face centres of the icosahedron. But is the resulting figure a cube? Consider a triad of three adjoining triangular faces, with the centres of the outer two triangles located at the endpoints of a cube edge. By symmetry of the icosahedron, such triads will be congruent wherever they are placed on the icosahedron. It follows that the edge lengths of the cube are all equal. Symmetry arguments also tell us that each cube edge is parallel to the base of the central triangle of its triad. So the 12 cube edges are parallel, in fours, to a set of three mutually perpendicular (and equal) edges of the icosahedron. Hence the cube edges are not only equal but meet at right angles as required. To inscribe a regular tetrahedron, we simply use our known result about inscribing a tetrahedron in a cube.

You might like to think about the number of different ways each of these solids can be inscribed in an icosahedron. We defer consideration of inscribing a dodecahedron inside an icosahedron until later when we look at duality. However, there is an easy way to place the icosahedron inside a cube, using those three pairs of equal, parallel and mutually orthogonal edges of the icosahedron: Golden rectangles With the dodecahedron and the icosahedron having so many regular pentagons associated with them, it shouldnt surprise us to find some golden rectangles occurring. We can show that the rectangles having as opposite edges those three pairs of equal, parallel and mutually orthogonal edges of the icosahedron, are in fact golden rectangles. Furthermore, the rectangles themselves are mutually perpendicular. Can we easily show that these rectangles are golden rectangles? It looks as though we might have to tackle the dreaded dihedral angles! Another idea is to work backwards, starting with three such golden rectangles of dimensions 2 x 2 .

Choosing our axes in the obvious way, this would mean that the vertices of our enclosing icosahedron (the vertices of the golden rectangles) would lie at ( polygons and polyhedra : paul scott : truncated icosahedron 2 )', where as usual the following prime denotes the cycling of the coordinate sets. This gives a total of (3 x 4) + (3 x 8) + (3 x 8) = 60 vertex sets as required. Occurrences in science and art The worlds of science and art seem to agree that the truncated icosahedron is one of the more interesting Archimedean solids. The oldest known picture of a truncated icosahedron is found in the book Libellus de quinque corpibus regularibus (On the Five Regular Bodies), by the Italian renaissance painter and mathematician Piero della Francesca, who lived 1420  1492. A copy of the volume can be found in the Vatican library.

The rendition of the truncated icosahedron at left comes from the book De Divina Proportione (On Divine Proportion) by Fra' Luca Pacioli, published around 1498. The drawing is ascribed to Leonardo da Vinci (1452-1519), who seems to have invented this type of framework model, which lets us view both the outside and the inside of the polyhedron. The concept of truncating a polyhedron was introduced by Johannes Kepler (1571-1630). Polyhedra were actually quite central to his research. In his work Prodromus Dissertationum Mathematicarum Continens Mysterium Cosmographicum (The Cosmographic Mystery) he argues that the distances of the planets from the sun are determined by the Platonic solids. If we think of a carbon atom being placed at each of the 60 vertices of a truncated icosahedron, we obtain the C60 molecule. You might lkike to learn more about Bucky balls , discovered in 1985.

The truncated icosahedron was also the configuration of the lenses used for focusing the explosive shock waves of the detonators in the Fat Man atomic bomb (see the Wikipedia reference below.) An example of the appearance of the truncated icosahedron in art is shown in this picture from an Italian cathedral. At the top we see an icosahedron. Truncating the vertices replaces each of them by a pentagonal face; it also converts each of the twenty former triangular faces into a hexagon. We can see the resulting truncated icosahedron at the bottom of the picture. The teddy bear figure at left was part of a 2006 UN exhibition in Sydney, Australia. References The Dodecahedron The dodecahedron has 30 edges, 20 vertices and 12 faces. Dodeca is a prefix meaning “twelve.” The dodecahedron is the only polyhedron I know of which is composed entirely of pentagons.

Buckminster Fuller describes what he calls a '6 pentagonal tensegrity sphere' in Synergetics I, 726.01, and which is shown at Marvin Solit's website at but I don't believe that structure can be built without tensegrity struts and tension wires. Figure 1 The dodecahedron is pentagonal both inside and out, as can be seen from Figure 1. Like the icosahedron, it has many golden section relationships, which we shall see. The dodecahedron is even more versatile then the icosahedron. The icosahedron contains and geometry, but the dodecahedron contains , Figure 2 Cube and tetrahedron in dodecahedron. Cube in gray, tetrahedron in green This view of the dodecahedron is significant in that it shows the 2 dimensional shadow of the decagon. The decagon itself is based upon the pentagon, the building block of the dodecahedron. See Pentagon and Decagon for more information.

Figure 2A -- Another view of tetrahedron (green) inside cube (blue) inside dodecahedron (orange) Figures 2 and 2A show how a cube and a tetrahedron can be placed inside a dodecahedron. These placements are 'nice' meaning that the vertices of the placed-in solids are all vertices of the dodecahedron. The cube, octahedron and tetrahedron are all based on root 2 and root 3 geometry: The relationship of the side of the cube to the radius of its enclosing sphere is r = sqrt(3) / 2. For the tetrahedron, The dodecahedron is capable of elegantly sustaining these and relationships, along with its own many relationships. Figure 3 -- octahedron inside tetrahedron Notice that the octahedron fits precisely on the bisected sides of the tetrahedron. The icosahedron cannot contain any of the other 5 solids 'nicely' on its vertices. The icosehedron and the dodecahedron are 'duals' (as are the cube and the octahedron).

By 'dual' is meant that if you put a vertex in the middle of all of the faces and connect the lines, you get the dual. By placing a vertex at the middle of all the faces of the dodecahedron you get an icosahedron, and vice-versa. Figure 4 shows the dual nature of the icosahedron and dodecahedron. Figure 4 Duals ---- dodecahedron inside icosahedron Notice that to create the dodecahedron, all we did was draw lines from each vertex of the icosahedron to every other vertex. The vertices of the dodecahedron are at the intersection points. We could just as easily have found the vertices of the dodecahedron by drawing lines on every triangular face of the icosahedron. Where those lines intersect is the center of the face, and a vertex of the dodecahedron. That occurs because the dodecahedron has 12 faces and the icosahedron has 12 vertices. Now for the standard analysis: What is the volume of the dodecahedron? We will use the pyramid method.

There are 12 pentagonal pyramids, 1 for each face, each pyramid beginning at O, the centroid. See Figure 1 and Figure 5. The volume of any n-sided pyramid is 1/3 * area of base * pyramid height. First we need to get the area of the base, which is the area of each pentagonal face: Figure 5 One pyramid on face BCHLG The area of the pentagon is the area of the 5 triangles which compose it. From Area of Pentagon we know This is approximately 1.720477401 . Now we need to find the height of the pyramid, OU. To do that, we need to find the distance from O to a vertex, lets say, OH. This distance will be the hypotenuse of the right triangle OUH. Since we already know UH, we can then get OU from the good ol’ Pythagorean Theorem. Imagine a sphere surrounding the dodecahedron and touching all of its vertices. OH is just the radius of the enclosing sphere. If you look at Figure 6, HOZ = GON = diameter. There is a line through HOZ to show the diameter.

Figure 6 Now look at the rectangle MIFK. The diagonal of it, MF, is also a diameter (MF = HZ). Notice that the long sides of the rectangle, MK, and IF, are diagonals of the two large pentagons ADMQK and FEIRP. We know from Composition of the Pentagon that the diagonal of a pentagon is * side of pentagon. We also can see from Figure 6 and more clearly in Figure 1 that the sides of the large pentagons themselves are diagonals of the pentagonal faces of the dodecahedron! (For instance, DA is a diagonal of the face ABCDE). That means each side of the large pentagons is * s and that MK (or any of the diagonals of a large pentagon) is * So MK = s. In fact, like the icosahedron, the dodecahedron is composed of rectangles divided in Extreme and Mean Ratio. In the icosahedron, we found these rectangles to be rectangles. In the dodecahedron, they are rectangles. In rectangle MKIF, MK = IF = , MI = KF = , as shown in figure 7.

Figure 7 -- showing vertices of the rectangle MKIF. There are 30 sides to the dodec, and therefore 15 different rectangles. All of this as explanation of finding the distance OH from Figure 5! Because we are not using trigonometry, we need OH in order to get the pyramid height, OU in Figure 5. Notice that MHFZ is also a rectangle and that HZ is the diagonal of it. If we can find HZ, then OH is just 1/2 of that. d² = HZ² = MZ² + HM² = diameter = HZ = Now we can find OU, the height of the pyramid. From Area of Pentagon we know the distance mid-face to any vertex of a pentagon = HOC = central angle = 41.81031488° Since each face of the dodecahedron is a pentagon, the surface angle = 108° While we're at it, lets get OX, the distance from the centroid to any mid- edge. What is the dihedral angle of the dodecahedron? Figure 9 AXH.

AH is one of the long sides of any of the 15 rectangles which compose the dodec. AX and HX are the height h of the pentagon. We know from Construction of the Pentagon Part 2 that the height h of the pentagon is: We know from Figure 7 that AH is , so IH = sin( IXH) = IH / XH = We recognize this ratio as our old friend the Phi Triangle with sides in ratio of So dihedral angle AXH = 116.5650512°. Distance from centroid to mid-face (h) = = 1.113516365s. Distance from centroid to mid-edge = = 1.309016995. Distance from centroid to vertex = = 1.401258539. Go back to Figure 6. We have colored the 4 internal pentagonal planes of the dodecahedron. U,X, W and V are the centers of these 4 planes which line up with the centroid O. What is the distance UX = WV? What is XW?

If we can find these out we can figure out more deeply how the dodecahedron is constructed. Figure 6, repeated In Figure 6 we can see that UH on the top plane is the distance from the pentagon center to a vertex. On plane ADMQK, XM is parallel to UH and is also the distance from that pentagon center to a vertex. UH is connected to XM by HM, a side of the pentagon face CDIMH. So we have a quadrilateral UHMX, with UH parallel to XM. From here we can derive UX, the distance between the two planes. Figure 10 -- dodecahedron planar distance. We know from Construction of the Pentagon Part 2 that the distance from the center of pentagon to a vertex = Therefore UH = The side of the large pentagon ADMQK in Figure 6 is, as we have seen, a diagonal of a dodecahedron face and so the side of the large pentagon = Therefore, XM = *UH, and XM is divided in Mean and Extreme Ratio at N (See Figure 10). = HM = s. Triangle MNH is right by construction.

So = HN = Notice: UH = UX. So the dodecahedron is designed such that the distance to the 2 large pentagonal planes from the top or bottom faces is exactly equal to the distance between the center and a vertex of any of the faces of the dodecahedron. This relationship is precisely what we saw in the icosahedron! That makes sense because the two are duals of each other. The difference is that the dodecahedron is entirely pentagonal, both internally, and externally, on its faces. What is the distance XW between the 2 large pentagonal planes ADMQK and FEIRP? Figure 6 is misleading, it looks like the distance must be MI or KF, the dodecahedron side, but it isn't. We already have enough information to establish this distance. UX = VW. UV = 2*height of any pyramid = So XW = UV - 2*UX = XW = UW is divided in Mean and Extreme Ratio at X. XV / WV = . XV is divided in Mean and Extreme Ratio at W.

Let's make a chart of these planar distances along the diameter of the dodecahedron as we did with the icosahedron: (see Figure 6): Relative Chart of Distances –– Pentagonal Planes of Dodecahedron Relative to the side of the dodecahedron (Available in the book) Here is a table of these relationships, letting UX = 1: (Available in the book) Note that the diameter of the enclosing sphere is HZ, not UV. We already know that the diameter is, from page 92, So what is the distance from U to the top of the sphere, and from V to the bottom of the sphere? Let T’ be the top of the sphere and B’ be the bottom of the sphere. Refer to Figure 6. If the radius is and the distance OU is , then Finally, let us demonstrate how the dodecahedron may be constructed from the interlocking vertices of 5 tetrahedron. We have already seen how the cube fits inside the dodecahedron, and how 2 interlocking tetrahedron may be formed from the diagonals of the cube.

As Buckminster Fuller has pointed out, however, the cube and the dodecahedron are structurally unsound unless bolstered by the additional struts supplied by the tetrahedron. Fuller concludes logically that the tetrahedron is the basic building block of Universe; yet it is the dodecahedron that provides the blueprint and forms the structure for the interlocking tetrahedrons. The dodecahedron unites the geometry of crystals and lattices (root 2 and root 3) with the geometry of Phi (root 5), found in the biology of organic life. The dodecahedron is entirely pentagonal, consisting of the geometry of Phi. Yet it contains the and geometry of the cube, tetrahedron, and octahedron. Remarkably, the sides of the cube are * side of the dodecahedron, because the cube side is the diagonal of a pentagonal face. Here is the key to the relationship of the first three Regular Solids and the much more complex icosahedron and dodecahedron.

Later on in this book we will discover a remarkable polyhedron that defines the relationship and provides the proper nesting for all 5 Platonic Solids, including the icosahedron, directly on its vertices. If a polyhedron could be called exciting, this one is IT! If you can’t wait, go to the last chapter of the book. Dodecahedron Reference Tables

Question: An icosahedron has how many faces?
Answer: 20